HDU_1009——老鼠的交易,性价比排序,最大化收益

Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

 

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

 

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

 

Sample Input

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

 

Sample Output

13.333
31.500

 /*
 算性价比j[i]/f[i],然后按性价比对每个房间的数据重新排序
 再累加最大化兑换比例的老鼠粮
 */
 #include <cstdio>
 #include <cstdlib>
 const int MAX = ;
 /*
 int compare(const void *a,const void *b)
 {
     //return *(double*)a - *(double*)b; //小——大
     return *(double*)b - *(double*)a;    //大——小
 }
 */
 int main()
 {
     int m,n,j[MAX]={},f[MAX]={};
     while(~scanf("%d%d",&m,&n))
         {
             if(m==- && n==-)
                 break;
             double j_f[MAX]={},temp=;
             for(int i=;i<=n;i++)
                 {
                     scanf("%d%d",&j[i],&f[i]);
                     j_f[i]=(double)j[i]/f[i];
                 }
             //快速排序：数组首地址，元素个数，一个元素大小，指向比较函数的指针
             //qsort(j_f+1,n,sizeof(double),compare);
             //好吧，写到一半发现排序不能这样用- -排序函数还是留着吧- -
             for(int i=;i<=n-;i++)
                 {
                     for(int k=i+;k<=n;k++)
                         {
                             if(j_f[i]<j_f[k])
                                 {
                                     temp=j_f[k];
                                     j_f[k]=j_f[i];
                                     j_f[i]=temp;

                                     temp=j[k];
                                     j[k]=j[i];
                                     j[i]=(int)temp;

                                     temp=f[k];
                                     f[k]=f[i];
                                     f[i]=(int)temp;
                                 }
                         }
                 }
             double ans=;
             for(int i=;i<=n;i++)
                 {
                     if(m>=f[i])
                         {
                             ans+=j[i];
                             m-=f[i];
                         }
                     else
                         {
                             ans+=m*j_f[i];
                             break;
                         }
                 }
             printf("%.3lf\n",ans);
         }
     return ;
 }